Step 1: Find the LCD. We factor each denominator completely.
\begin{align*}
x^2-2x \amp = x(x-2)\\
x^2-4 \amp = (x-2)(x+2)
\end{align*}
The LCD is \(~x(x-2)(x+2)\text{.}\)
Step 2: We build each fraction to an equivalent one with the LCD as its denominator. The building factor for \(\dfrac{x-4}{x(x-2)}\) is \(\alert{(x+2)}\text{.}\) We multiply the numerator and denominator of the first fraction by \(\alert{(x+2)}\text{:}\)
\begin{equation*}
\dfrac{x-4}{x(x-2)} = \dfrac{(x-4)\alert{(x+2)}}{x(x-2)\alert{(x+2)}} = \dfrac{x^2-2x+8}{x(x-2)(x+2)}
\end{equation*}
The building factor for \(\dfrac{4}{(x-2)(x+2)}\) is \(\alert{x}\text{.}\)
\begin{equation*}
\dfrac{4}{(x-2)(x+2)} = \dfrac{4\alert{x}}{(x-2)(x+2)\alert{x}}
\end{equation*}
Step 3: The fractions are now like fractions, so we add them by combining their numerators.
\begin{align*}
\dfrac{x-4}{x^2-2x} + \dfrac{4}{x^2-4} \amp = \dfrac{x^2-2x+8}{x(x-2)(x+2)} + \dfrac{4x}{x(x-2)(x+2)}\\
\amp = \dfrac{x^2+2x+8}{x(x-2)(x+2)}
\end{align*}
Step 4: Finally, we reduce the fraction.
\begin{equation*}
\dfrac{x^2+2x+8}{x(x-2)(x+2)} = \dfrac{(x+4)\cancel{(x-2)}}{x\cancel{(x-2)}(x+2)} = \dfrac{x+4}{x(x+2)}
\end{equation*}